给定以下接口：
type A<'t> =
abstract Scratch: 't
如何简单地创建这样的界面：
type X<A<'t>> =
abstract f: A<'t>
我也尝试过这个：
type X<'t,'y when 'y:A<'t>> =
abstract f: 'y
还有这个：
type X<'t,A<'t>> =
abstract f: A<'t>
但没有一个对我有用。
如果您希望将其作为界面，请尝试以下操作：
type IA<'a> =
abstract Scratch: 'a
type IX<'a, 'b when 'a :> IA<'b>> =
abstract F: 'a
type RealA() =
interface IA<int> with
member __.Scratch = 1
type RealX =
interface IX<RealA, int> with
member __.F = RealA()
如果要抽象类：
[<AbstractClass>]
type A<'a>() =
abstract member Scratch: 'a
[<AbstractClass>]
type X<'a, 'b when 'a :> A<'b>>() =
abstract F: 'a
type RealA'() =
inherit A<int>()
override __.Scratch = 1
type RealX'() =
inherit X<RealA', int>()
override __.F = RealA'()
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